是的,您的方法是正确的!
我在下面发布以下代码:
%Rotating Pendulum Sym Main
clc
clear all;
%Define parameters
global M K L g C;
M = 1;
K = 25.6;
L = 1;
C = 1;
g = 9.8;
% define initial values for theta, thetad, del, deld
e_0 = 1;
ed_0 = 0;
theta_0 = 0;
thetad_0 = .5;
initialValues = [e_0, ed_0, theta_0, thetad_0];
% Set a timespan
t_initial = 0;
t_final = 36;
dt = .01;
N = (t_final - t_initial)/dt;
timeSpan = linspace(t_final, t_initial, N);
% Run ode45 to get z (theta, thetad, del, deld)
[t, z] = ode45(@RotSpngHndl, timeSpan, initialValues);
%initialize variables
e = zeros(N,1);
ed = zeros(N,1);
theta = zeros(N,1);
thetad = zeros(N,1);
T = zeros(N,1);
V = zeros(N,1);
x = zeros(N,1);
y = zeros(N,1);
for i = 1:N
e(i) = z(i, 1);
ed(i) = z(i, 2);
theta(i) = z(i, 3);
thetad(i) = z(i, 4);
T(i) = .5*M*(ed(i)^2 + (1/3)*L^2*C*sin(theta(i)) + (1/3)*L^2*thetad(i)^2 - L*ed(i)*thetad(i)*sin(theta(i)));
V(i) = -M*g*(e(i) + .5*L*cos(theta(i)));
E(i) = T(i) + V(i);
end
figure(1)
plot(t, T,'r');
hold on;
plot(t, V,'b');
plot(t,E,'y');
title('Energy');
xlabel('time(sec)');
legend('Kinetic Energy', 'Potential Energy', 'Total Energy');
这里是 ode45 的函数句柄文件:
function dz = RotSpngHndl(~, z)
% Define Global Parameters
global M K L g C
A = [1, -.5*L*sin(z(3));
-.5*L*sin(z(3)), (1/3)*L^2];
b = [.5*L*z(4)^2*cos(z(3)) - (K/M)*z(1) + g;
(1/3)*L^2*C*cos(z(3)) + .5*g*L*sin(z(3))];
X = A\b;
% return column vector [ed; edd; ed; edd]
dz = [z(2);
X(1);
z(4);
X(2)];